Lesson video

Hello there, I'm Mr. Forbes and welcome to this lesson from the Energy of Moving Particles unit.

The lesson's called calculating specific heat capacity.

By the end of this lesson, you're going to be able to carry out a wide range of calculations involving the equation, energy change is equal to mass times specific heat capacity times temperature change.

There are just two key phrases for this lesson.

The first is specific heat capacity and that's the change in internal energy and the temperature of one kilogramme of a material changes by one degree Celsius.

The second is joules per kilogramme per degree Celsius, and that's the unit per specific heat capacity of a material.

This lesson's in three parts and the first part we're going to have a look at the factors that affect the heating of an object and define specific heat capacity clearly.

In the second part of the lesson, we're going to be looking at a range of specific heat capacity calculations.

And in the third and final part, we're gonna be trying to solve some more complex problems involving specific heat capacity.

So let's get on by defining what specific heat capacity is.

You should already realise that it takes longer to boil a larger amount of water than a smaller amount of water.

So if you've got an electric kettle and you fill it to the top, it's gonna take longer to boil than if you only half fill that kettle.

So for example, if I've got a kettle and it's got 0.

2 kilogrammes of water, it might boil in 30 seconds.

But if I double that amount of water to 0.

4 kilogrammes, it's going to take 60 seconds for the same kettle to boil that water.

And if I double it again, again, the time doubles.

So doubling the mass of water, doubles the amount of time taken to boil it in that kettle.

The kettle's providing the same amount of energy every second to the water.

So it's taking twice as much energy to boil twice as much water.

That means the amount of energy needed to increase the temperature of a substance is directly proportional to its mass.

Let's check if you understand that idea.

I've got three pans of water here, they're being heated by an identical heater from below and they're all starting at the same temperature, room temperature, which pan is going to reach boiling point first.

So look carefully at the three diagrams, make that decision, pause the video, make a decision, and restart, please.

Welcome back.

Hopefully you selected B, B's got the smallest amount of water in it, the lowest mass.

So it's going to be easier to boil that.

Well done if you selected it.

The longer that you actually heat a substance, the more energy you're going to give it and the greater it's change in temperature.

So, for example, if I've got two objects here, both starting at the same temperature and they're both identical, they're one kilogramme chunks of copper and I heat the first one for 100 seconds, its temperature has gone up to 50 degrees Celsius.

So the change in temperature there is 30 degrees.

For the second one, let's say I heat it for twice as long, so 200 seconds of heating and its end temperature actually is 80 degrees Celsius.

It's gone up by 60 degrees C.

So what you should be able to see from that is doubling the energy provided, doubles the temperature change of an object, the change in temperature is directly proportional to the energy provided to it.

Let's see if you understand that second concept.

I've got three pans each of them this time containing one kilogramme water, so the same amount of water in each pan.

They're again being heated by identical heaters, but this time for different length of time.

And you can see the length of time there, one minute, three minutes, five minutes.

Which water will have the greatest temperature rise? Pause the video, make your decision and restart please.

Welcome back.

Hopefully you selected the third one, option C, that's five minutes of heating.

I'm gonna be providing you with more energy, so it's gonna have a greater temperature rise.

Well done if you've got that.

If you heat different substances then the temperature changes won't necessarily be the same, even if they've got the same mass and same starting temperature.

So I've got two different substances here, I've got iron and copper.

If I start with a temperature of 20 degrees Celsius and heat both the same amount, an equal amount of heating for the same amount of time, then the temperature change isn't the same for those two.

The temperature change of the iron might be up to 72 degrees Celsius, and the temperature change of the copper would only go up to 60 degrees Celsius.

So the substances aren't heating up the same way and that's because they have different, specific heat capacities.

And the specific heat capacities related to the temperature change of an object and the energy provided to it.

So the energy needed to increase the temperature for the different metals by the same amount is directly proportional to its specific heat capacity of property of the material that's different for different substances.

So I've got three pans again, these pans contain one kilogramme of liquids with different specific heat capacities and they're warmed again by identical heaters starting from room temperature for one minute each.

So at the same time, the same starting temperature and the final temperatures are then measured and they're shown in the diagram here.

Which of those liquids needed the least energy to heat per degree Celsius> So pause the video, make your decision and restart, please.

welcome back.

Hopefully you selected option C.

The reason for that is that's got the largest temperature increase, so it needed the least amount of energy to heat it by one degree Celsius.

Remember all three of those were given the same amount of energy and that one increased its temperature the most.

So well done if you selected that.

Now, let's look at that definition of specific heat capacity.

The property of a material that tells you how difficult it is to change its temperature.

The heat capacity of a material is the internal energy change and the temperature of one kilogramme of the material changes by one degree Celsius.

So it's a measure of how difficult it is to change the temperature of one kilogramme by one degree.

Some examples, we've got water, that's got a specific heat capacity of 4,000 joules per kilogramme per degree Celsius.

So it takes 4,000 joules to heat one kilogramme of it by one degree Celsius.

And if I wanted to heat two kilogrammes of it by one degree then I'd need twice as much energy, 8,400 joules.

Iron's got a different specific heat capacity, it's much easier to heat to iron.

It's only 110 joules per kilogramme per degree Celsius.

So it takes 110 joules to heat one kilogramme by one degree C.

And if I wanted to increase its temperature by twice as much by two degrees C, it would take twice as much energy.

As we've seen in the early part of the lesson, the energy required to change the temperature of an object depends on three factors.

It's directly proportional to the mass.

We need a large energy change to heat large objects, objects with more mass.

It's also directly proportional to the change in temperature.

We need a larger energy change to give a larger temperature change.

And finally, the specific heat capacity.

Some materials are easier to heat up than others, and the energy required is directly proportional to that specific heat capacity as well.

All those three factors can be linked together in a single equation.

So here is that equation linking all the factors together and it is energy change is equal to mass times specific heat capacity times temperature change.

And if you write there in symbols, we have this delta E equals MC delta theta, where delta E is the energy change in joules, M is the mass in kilogrammes, and C is the specific capacity in joules per kilogramme per degree Celsius, delta theta is the temperature change in degrees celsius.

Let's have a look at an example of using that equation then.

So I've got a question here.

Iron's got a specific heat capacity of 450 joules per kilogramme per degree Celsius.

How much energy is required to increase the temperature of two kilogramme block of the iron by 30 degrees Celsius? So what we do is we write out the equation, energy change equals MC delta theta, we substitute in the values from the question.

So the mass was two kilogrammes, the specific heat capacity was 450 joules per kilogramme per degree Celsius and the temperature change is 30 degrees Celsius.

And then we can just do the calculation, giving us an answer of 2.

7 kilojoules when we've multiplied all of those out.

I'll try another one and then you can have a go.

So I've got glass that's got a specific heat capacity of 840 joules per kilogramme per degrees Celsius.

Calculate the energy required to increase the temperature of a 1.

5 kilogramme block by 80 degrees Celsius.

So again, I write out the equation, I substitute in the values from the question carefully and I perform that calculation given me 101 kilojoules.

Now, it's your turn.

What I'd like you to do is to calculate the energy change in the situation shown here.

So pause the video, calculate the energy change and restart, please.

Welcome back.

Hopefully, you performed a calculation that looked something like this.

Write out the equation, substitute the values and the questions carefully, the mass, specific capacity and temperature change, multiply all those out and you've got an energy change of 235 kilojoules.

Well done if you've got that.

Now, it's time for the first task of the lesson and what I'd like you to do is answer these two questions.

So the first one is about heating the liquid and the second one, heating different materials and using the specific heat capacity equation.

So pause the video, work out your answers to those and restart please.

Welcome back, here are the solutions.

So the temperature rise should have been 60 degrees Celsius.

The reason for that is the material's being provided with six times the original energy.

So the temperature rise would be six times as great, but the mass is three times as large.

So that would only give a third of the temperature rise.

Combining those two factors together, the temperature rise is double.

The reason you might not get an exact answer is because some of the liquid might evaporate, so energy would be dissipated or lost to the surroundings.

And if you perform the calculations, you should have got energy values as shown in the table though.

Well done if you've got any of those.

And now it's time for the second part of the lesson and then we're going to be using that specific heat capacity equation, rearranging it to solve a range of problems. So as we saw earlier, the equation linking energy change, mass, specific capacity, and temperature change is shown here.

There's four variables in that equation, and that means it's going to be a little bit more difficult to rearrange compared to other simpler equations that only have three.

So the approach we're gonna take in this lesson to finding values is we're going to be substituting the values into the equation, simplifying what we know some of the data and then that will make finding the solution a little bit easier.

So I'll show you some examples of how to do that here.

Let's have a look at an example of finding mass.

So I've got a question here.

The specific capacity of water is 4,200 joules per kilogramme per degree Celsius.

A kettle provides 189 kilojoules of energy when heating the water from 15 degrees to 90 degrees, calculate the mass of water in the kettle.

So what we do is we write up the equation, we've got energy changes, mass times specific capacity times change in temperature, and it put in the values.

So I'm gonna put in the value for the energy change.

It says 189 kilojoules, so it's 189,000 joules, I don't know the mass, so I'll leave that as a symbol M, then I've got a specific capacity, and then I've got the change in temperature.

And you'll see though I've had to calculate the change in temperature, it's gone up to 90 degrees, but it started at 15 degrees Celsius.

So let's simplify that a little bit first by working out that change in temperature and it's 75 degrees Celsius, and the next stage is I can just multiply the 4,200 times that 75 degrees Celsius and that gives me a value of 315,000.

And now I've got a much simpler equation.

I've got 189,000 joules is equal to mass times 315,000 joules per kilogramme.

So I can rearrange that as this only three values now.

So I'm gonna rearrange it in terms of M.

So the mass is equal to the energy divided by that 315,000 joules per kilogramme and I can just simply solve that using a calculator that gives me a mass of 0.

6 kilogrammes.

Okay, let's see if you can repeat that process.

I've got a specific heat capacity of lead here, 130 joules per kilogramme per degree Celsius.

I've got a heater providing 8.

9 kilojoules of energy heating that block of lead from 20 to 60 degrees.

Calculate the mass of the lead block.

So pause the video, perform the calculation I did before and restart, please.

Welcome back.

Hopefully you selected 1.

54 kilogrammes and here's the mathematics behind that.

We've got the equation written out, substitute the values, then I found the change in temperature being 40 degrees Celsius.

I've calculated the 130 joules per kilogramme per degree Celsius times a 40 degrees Celsius to give me a simpler version of the equation.

Then I've rearrange it in terms of mass.

So we've got mass is 8,000 joules divided by five, 5,200 joules per kilogramme, and that gives me a final mass of 1.

54 kilogrammes.

Well done if you've got that.

We can do a similar thing if we're trying to find a change in temperature.

So here's another example.

I've got a Bunsen burner providing 3.

6 kilojoules of energy to a 0.

20 kilogramme sample of olive oil, and that's got a specific heat capacity of 1,980 joules per kilogramme per degree Celsius.

And I'm going to find the change in temperature for that oil.

So as before I write out the equation and I substitute in the values that I know.

So I've got my energy change written here, 3,600 joules, I've got my specific capacity, and I've got my mass as well.

What I haven't got is the change in temperature.

So I leave that as delta theta, the next thing I do is I perform all the parts of the calculation I can do with the numbers I've got.

So I'm the 0.

20 by 1,980 and that gives me this stage of the equation.

And now I can do the rearrangement.

I can write out delta theta is equal to 3,600 joules divided by 396 joules per kilogramme.

And that gives me a temperature change of 9.

1 degrees Celsius.

Here's one of those calculations for you to do.

What I'd like you to do is to pause the video, read through the information and work out your answer, and then restart, please.

Welcome back, hopefully you selected answer B, 8.

0 degree Celsius.

We show you the mathematics here.

We've got the energy change equation.

We substitute in the values, we simplify in terms of delta theta and then so we get delta theta is 8,400 joules divided by 1050 joules per kilogramme.

And that gives me a final temperature change of 8.

0 degrees Celsius.

Well done if you got that.

And the final way to use the equation is to actually find a specific heat capacity based upon all the other data.

So I've got another question here.

An engineer is testing properties of a metal sample.

They use an electric heater to provide 0.

50 kilogramme sample of the metal with 8.

0 kilo joules of energy.

And its temperature increased from 24 degrees to 62 degrees Celsius.

Calculate the specific heat capacity.

Whereas before we write out the equation, energy change MC delta theta, we substitute in the values that we can see in the question.

8,000 joules, 0.

5 kilogrammes and the temperature change there from 24 degrees up to 62.

So we need the difference of those two values.

So we'll calculate that, and that was 38 degrees Celsius.

So we're left with this stage of the equation and then we can multiply the parts that we've got there, the 0.

5 kilogrammes times a 38 degrees Celsius to give us this version and finally rearrange that in terms of C.

So we get C equals specific heat capacity equals 8,000 joules divided by 19 kilogrammes per degree Celsius.

And that gives us a specific heat capacity of 421 joules per kilogramme per degree Celsius.

So it's your go again, I'd like you to read the information here about a rock sample being heated and find the specific heat capacity of the rock bleeders.

Pause the video, find that specific heat capacity and restart.

Welcome back.

Hopefully you selected option D, 972 joules per kilogramme per degree Celsius.

And again, I'll show you all the mathematical stages here which substitute in all the values and that simplifies down to C equals 14,000 joules divided by 14.

4 kilogrammes per degree Celsius.

And that gives that final of 972 joules per kilogramme Celsius.

Well done if you got that right.

Okay, we've reached the second task of the lesson and I've got two questions for you to answer here.

Both of them involving use and manipulation of that specific heat capacity equation.

So what I'd like you to do is to pause the video, read through those questions and answer them, and then restart when you're done, please.

Welcome back.

And for the first part of the question we've got question one here.

Part A, calculate the energy provided to the water while the water was being given 8,000 joules of energy per second for 60 seconds for one minute.

So that gives us an energy value of 480,000 joules.

Then to find the final temperature of the water or what we can do is use the specific heat capacity equation substituting the values, including that value for the energy we just calculated.

That gives us a temperature change of 76 degrees Celsius.

But remember, we've been asked to find a final temperature of the water.

So what we can do though is add the initial temperature to that and that gives us a final temperature of 96 degrees Celsius just short of boiling.

Well done if you've got that.

And here's the answer to question two completed table.

And I've shown you the example calculation for gold here.

The other calculations would be very similar.

We substitute in the values we know, which is the energy, the mass, and we find that temperature change based upon the start and end temperatures and that gives us an equation for C and for gold it was 8,000 joules divided by 60.

2 kilogrammes per degree Celsius.

And that gives us a final specific capacity for gold of 133 joules per kilogramme per degree Celsius.

The other calculations should look very similar.

Well done if you've got that table right.

And now it's time to move on to the third and final part of the lesson.

And in it we're going to look at some more complex problems involving specific heat capacity.

Let's do that.

Many of the calculations that involve specific heat capacity are more than one part and we're going to look at questions that involve two or three steps in order to find final solutions.

So I've got an example over the next couple of slides and we'll break it down into the different parts we need to do.

So here's our first example.

I've got a blacksmith working with a 2.

5 kilogramme piece of steel at a temperature of 750 degrees Celsius.

So quite hot.

To cool it quickly, they're gonna plunge it into a tank of water, 20 kilogrammes of water at a temperature of 20 degrees Celsius.

The temperature of the water rises to 50 degrees Celsius and the specific heat capacity of water is 4,200 joules per kilogramme per degree Celsius.

Calculate the specific heat capacity of the steel.

So to answer that, we need to have two steps.

We need to find the energy transferred into the water from the steel and then we need to use that value to find the specific heat capacity of the steel.

So let's look at those stages.

So the question's repeated there and step one, as I said, is we're going to find the energy change for the water that's going to allow us to find things about the steel a bit later on.

So the energy change for the water first, write out the equation, energy change, mass specific heat capacity times temperature change.

And remember, we're dealing with the water, so we use the values for the water here.

There's 20 kilogrammes of water, it's specific heat capacity is 4,200.

And the temperature change of the water, well, it's gone up from 20 degrees Celsius to 50 degrees Celsius.

So that's a temperature change of 30 degrees and that will allow us to find the energy change of the water.

Multiplying all those two together, we've got an energy change of 2,520,000 joules.

Okay, before I go onto the next part of the question, I'd like you to think about something.

I've got a hot piece of steel and it's dropped into water, and it's cooled.

The internal energy of the water increases by 2000 kilojoules.

How much has the internal energy of the steel decreased by? So pause the video, have a think about that, Select that answer and restart.

Welcome back.

Hopefully you selected answer B, 2000 kilojoules.

That's because of the principle of conservation of energy.

The internal energy of the water has increased by 2000 kilojoules.

So there must have been a loss of 2000 kilojoules by the steel.

Well done if you selected that.

So we're gonna continue with our example of the blacksmith.

We've got the energy change for the steel now.

So that's my equation for the energy change for the steel.

And I know that the energy change of the water was 2,520,000 joules, and that's the same as the energy change for the steel because of that conservation of energy.

I've got the mass of the steel in there, and I've now got the temperature change of the steel.

The steel cooled from 750 degrees to 20 degrees.

So I've got the temperature change and I'm just gonna find that value for C.

So I'll do mathematics to find the temperature change.

It's 730 degrees, and now I can simplify that by multiplying the 2.

5 kilogrammes by the 730 degrees and that can be this version, and the next stage I can rearrange that equation And finally I can do that calculation and that gives me the specific heat capacity of the steel.

Let's have a look at another example.

I've got a central heating boiler and it can transfer 18 kilojoules of energy per second to heat water for a bath.

A total of 12 kilogrammes of water at a temperature of 15 degrees Celsius passes through the heater each minute.

What temperature does the water leave the heater at? So to answer this question, I need three steps.

The first step is to find the amount of water passing through the heater each second.

The second step is to find the increase in water temperature using the energy provided per second.

And finally to find the new temperature of the water.

Let's see if you can do that first step.

I've got a total of 15 kilogrammes of water at a temperature of 20 degrees Celsius passing through a pipe in a minute.

How much water is passing through per second? Pause the video, make your selection and restart, please.

welcome back.

Hopefully you selected option A, 0.

25 kilogrammes.

Water passing through per second is 15 kilogrammes in a minute.

So we divide that by the 60 seconds to get per second.

And that's 0.

25.

Well done if you've got that.

So back to my question, let's calculate the water passing through per second.

For this scenario, we've got 12 kilogrammes passing through per minute, so that's 0.

2 kilogrammes per second.

So now I can find the temperature change of that water.

I've got the heater providing O.

2 kilogrammes of water with 18 kilojoules of energy every second.

So I write up my equation for energy change and I substitute my values, I've got a massive 0.

2 kilogrammes, I've got a specific heat capacity of 4,200 joules per kilogramme per degree Celsius.

And I know the energy change in the water is 18,000 joules, 18 kilojoules.

So I'm just looking for Delta theta.

So I follow the same procedures as I've done before, multiplying the values that I can, then simplifying to get an equation for delta theta, and then I'll find that that temperature change is 21.

4 degrees Celsius.

But I was actually asked what temperature the water leave at.

So I need my new water temperature and it went in at 15 degrees.

So what I need to do is to add those values together.

The new water temperature is the original temperature plus the increase and that gives me a final value of 36.

4 degrees Celsius.

So very hot water there.

A question for you now, Which of these boilers would be able to increase the temperature of the water by the greatest amount? So have a read through the data we're provided there and make a decision.

Pause the video, select A, B, or C, and then restart, please.

Welcome back.

Hopefully you selected option C, the eight kilowatt boiler providing energy just to 0.

3 kilogrammes of water each second.

That's got the highest heating effect per kilogramme of water.

Well done if you've got that.

Okay, it's time for the final task of the lesson.

And I've got two questions involving manipulation of this specific, heat capacity equation in multiple steps.

So what I'd like you to do is to read through those, answer those questions for me, please.

So pause video, try that, and restart.

Welcome back.

Let's have a look at the solutions to those.

So for the first one, I need to find the energy provided to the oil first.

So the energy provided to the oil is given there, it's 56,000 joules and then I can find the energy per second.

Five minutes is 300 seconds.

I've got the total energy provided.

Then divide that by 300 seconds and that's 187 joules per second.

Well done if got that.

And for the second one, we're calculating the internal energy change for the water first.

So the energy change equation is there.

I've put in the mass of water, specific heat capacity and the temperature change, and that's 840 joules.

And then I can calculate the specific heat capacity of the stone using the same equation.

But this time, I'm looking for C for the stone.

So I put in the values, simplify, get the relationship for C, and that gives me specific capacity of 1,680 joules per kilogramme per degree Celsius.

Well done if you've got that.

And now we've reached the end of the lesson.

So a quick summary.

The specific heat capacity of a material is the change in internal energy or the temperature of one kilogrammes of material changes by one degree Celsius.

Energy change is mass times specific heat capacity times temperature change.

And that's shown as symbols here as well with all the definitions below.

Well done for reaching the end of the lesson.

I will see you in the next one.